Difference between revisions of "2015 AIME I Problems/Problem 3"
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==Solution 4== | ==Solution 4== | ||
− | Notice that <math>16p+1</math> must be in the form <math>(a+1)^3 = a^3 + 3a^2 + 3a + 1</math>. Thus <math>16p = a^3 + 3a^2 + 3a</math>, or <math>16p = a\cdot (a^2 + 3a + 3)</math>. Since <math>p</math> must be prime, we either have <math>p = a</math> or <math>a = 16</math>. Upon further inspection and/or using the quadratic formula | + | Notice that <math>16p+1</math> must be in the form <math>(a+1)^3 = a^3 + 3a^2 + 3a + 1</math>. Thus <math>16p = a^3 + 3a^2 + 3a</math>, or <math>16p = a\cdot (a^2 + 3a + 3)</math>. Since <math>p</math> must be prime, we either have <math>p = a</math> or <math>a = 16</math>. Upon further inspection and/or using the quadratic formula, we can deduce <math>p \neq a</math>. Thus we have <math>a = 16</math>, and <math>p = 16^2 + 3\cdot 16 + 3 = \boxed{307}</math>. |
== See also == | == See also == |
Revision as of 13:11, 3 March 2020
Problem
There is a prime number such that
is the cube of a positive integer. Find
.
Solution 1
Let the positive integer mentioned be , so that
. Note that
must be odd, because
is odd.
Rearrange this expression and factor the left side (this factoring can be done using or synthetic divison once it is realized that
is a root):
Because is odd,
is even and
is odd. If
is odd,
must be some multiple of
. However, for
to be any multiple of
other than
would mean
is not a prime. Therefore,
and
.
Then our other factor, , is the prime
:
Solution 2
Since is odd, let
. Therefore,
. From this, we get
. We know
is a prime number and it is not an even number. Since
is an odd number, we know that
.
Therefore, .
Solution 3
Let . Realize that
congruent to
, so let
. Expansion, then division by 4, gets
. Clearly
for some
. Substitution and another division by 4 gets
. Since
is prime and there is a factor of
in the LHS,
. Therefore,
.
Solution 4
Notice that must be in the form
. Thus
, or
. Since
must be prime, we either have
or
. Upon further inspection and/or using the quadratic formula, we can deduce
. Thus we have
, and
.
See also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.